Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}3x+y &= 8 \\ -2x-y &= -9\end{align*}$
Solution: Begin by moving the $y$ -term in the second equation to the right side of the equation. $-2x = y-9$ Divide both sides by $-2$ to isolate $x$ $x = {-\dfrac{1}{2}y + \dfrac{9}{2}}$ Substitute this expression for $x$ in the first equation. $3({-\dfrac{1}{2}y + \dfrac{9}{2}}) + y = 8$ $-\dfrac{3}{2}y + \dfrac{27}{2} + y = 8$ Simplify by combining terms, then solve for $y$ $-\dfrac{1}{2}y + \dfrac{27}{2} = 8$ $-\dfrac{1}{2}y = -\dfrac{11}{2}$ $y = 11$ Substitute $11$ for $y$ in the top equation. $3x+ 11 = 8$ $3x+11 = 8$ $3x = -3$ $x = -1$ The solution is $\enspace x = -1, \enspace y = 11$.